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AY 20: Basic Astronomy and the Galaxy
Solution Set 3
November 6, 2009
Swarnima Manohar
[email protected]
1. C&O Problem 7.3
(a) An eclipse will just occur when the common tangent is along the line of sight, as shown in the figure. In this case,
Sin90 °  i  Cosi 
r1
x

r2
ax

r1  r2
a  x  x

r1  r2
a
The minimum i at which an eclipse occurs is then
i  ArcCos
r1  r2
a

(b)
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i  ArcCos
11 Rsun
2 AU

180

. Rsun  6.96 1010 , AU  1.496 1013 
i  88.5337
Thus, i = 88.5°.
2. C&O Problem 7.6
(a)
mA

mB
vBr

22.4
 4.14
5.4
vAr
(b) Using Eq. 7.6 in Carroll and Ostlie,
m A  mB 
P
vAr  vBr 3
2G
Sini3
g .
P  6.31  365.25  24  3600, G  6.67 108 , vAr  5.4 105 , vBr  22.4 105 , i    2
mA  mB  1.02085  1034 g
which is 5.13 MŸ .
(c)
mA
Solve
 4.14, mA  mB  5.13 Msun , mA, mB
mA  0.  4.13195 Msun , mB  0.  0.998054 Msun 
mB
m A = 4.13 MŸ , mB = 0.998 MŸ .
(d) Assuming the orbital separation is much larger than the stellar radii, and that the orbits are circular, we can treat the velocity
of the stars during eclipse as completely in the plane of the sky. For circular orbits, the maximum radial velocities given are the
constant velocities throughout the orbit. The relative velocity is then in km/s
v  5.4  22.4
v  27.8
It takes a time tb - ta for the smaller disk to enter the larger one, so the radius of the smaller star must be
rs 
1
2
v tab  cm . v  27.8 105 , tab  0.58  24  3600 
rs  6.96557  1010 cm
which is about 1 RŸ . (Note that assuming iº 90° as we do here implies the transit is across a diameter of the large disk.) Now
consider the point on the smaller disk that first eclipses the larger. From time ta to tc this point traces the diameter of the larger
disk (see Fig. 7.9 in Carroll and Ostlie), so the radius of the larger disk is
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rl 
1
2
v tcb  tab  cm . v  27.8 105 , tab  0.58  24  3600 , tcb  0.64  24  3600 
rl  1.46517  1011 cm
which is 2.11 RŸ .
(d) Following example 7.3.2 in Carroll and Ostlie, the ratio of fluxes between the primary minimum and maximum light is
B p  B0 = 1005.40-9.205 = 0.030, and the ratio of fluxes between the secondary minimum and maximum light is Bs  B0 =
1005.40-5.445 = 0.964. Then
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Ts
1  B p  B0
Tl
Ts
14
1  Bs  B0

. Bp  B0  0.030, Bs  B0  0.964
 2.27833
Tl
The derivation in the text assumes that the smaller star is hotter, i.e. that the primary eclipse is when the smaller star passes
behind the larger. Can we back this up with the data? Assuming this is true, then in the primary eclipse we see only the larger
star, which gives 100m0 -m p 5 = 1005.40-9.205 = 3.02% of the total brightness. In the secondary, we then expect to see the remaining 96.98% of the flux from the smaller star, plus the fraction rs  rl 2 = 0.225 (from part c) of the big star’s 3.02% that is unob-
structed, for a total of 97.7%. We observe 100m0 -ms 5 = 1005.40-5.445 = 96.38%, which is close considering we only two significant digits in the times.
If we instead imagine that the larger star is hotter, so that the primary eclipse occurs when the smaller star is in front, we find that
the 96.38% of light visible in the secondary is due to the larger star, and that during the primary we expect 0.225 of this (plus the
3.62% contribution of the smaller star), which is far in excess of the 3.02% observed.
Another way to estimate the temperature ratio is to note that the primary minimum is 0.030 of the maximum brightness, so that
0.030 T4s Rs2  T4l Rl2   T4l Rl2

Ts
1

1
0.030
Tl
Rl
2 14
 3.5
Rs
using the radii from (d). This does not agree particularly well with the 2.3 we found above, indicating that the timing data in this
problem are not consistent with the photometry under the assumption that the smaller star passes directly across a diameter of the
larger star’s disk. For this homework either answer is acceptable.
3. C&O Problem 7.11
For simplicity, assume circular orbits. (In any case, eccentricity can be deduced from distortions of
the radial velocity curve from sinusoidal.) Denoting the masses of the planet and star by M p and Ms , Kepler’s third law says
P2 
4 2
G Mp  Ms 
a3 
4 2
a3
G Ms
This is because M p ` Ms , so
a
G Ms P2
13
4 2
The orbital velocity of the planet is approximately
vp 
2a
P

2  G Ms
13
P
and that of the star is reduced approximately by the mass ratio:
vs 
Mp
vp
Ms
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We can measure the period P and estimate Ms by determining its spectral type with our spectrum. Thus, we know v p . If we could
measure vs , then the above equations would allow us to solve for M p . However, what we measure is vrs = vs sin i, the radial
lower limit on M p = Ms vs  v p , but cannot say more.
component of the star’s velocity, and the orbital inclination i is in general unknown. Since vrs is a lower limit on vs , we get a
4. C&O Problem 7.13
Under the assumptions of a uniform solar disk and no luminosity from Jupiter, the decrease is just given by the fraction of the
solar disk blocked by Jupiter:
 RJ2
 Rü2
RJ

7.1 109 cm
2

7 1010 cm
Rü
2
 1.1 
5. C&O Problem 9.20
For a plane grey atmosphere in LTE under the Eddington approximation, Carroll and Ostlie derive
4
3
I  Frad  
2
3
in Eqn 9.50. Using Frad = sT4e (by definition) and I = S = B = sT4 /p (in LTE), this becomes
T4 
3
4
T4e  
2
3
From this we see that T = Te at t = 2/3. (Note that along any line of sight, you are seeing photons from an average t = 1; however, when averaged over angles, this corresponds to an average vertical depth of tv =2/3.) This can also be derived by assuming a
liinear source function S = a +bt, which is another form of the Eddington approximation.
6. C&O Problem 10.11
The goal is to approximate the expression (eqn 10.62)
1
44.027 T8
3   50.9 2 Y3 T3
8 f3  
as a power law of the form:
3   T8
near T = 108 T8 = 1. We equate the two equations, take log of both sides, require their log-derivatives to match, and plug in T8 =
1 to solve for a:

 ln T8
44.027 T8
ln T8   const  ln T3

8 
  3 
1
44.027
 41
T8
7. C&O Problem 10.12
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The mass of the proton
1
1
H 1.0073 u, and that of
4
2
He is 4.0026 u. In the text, the masses of
2.0141 u and 3.0160 u, but note that the mass quoted here for
2
1
2
1
H and
3
2
He are also given as
H seems to
+
include the mass of the positron e , based on comparison with other references. To compute the Q value we subtract the total
mass of the products from that of the reactants (with heat released then corresponding to Q > 0):
Reaction
Q
10.37
2  1.0073  2.0141 u  0.0005 u  0.5 MeV
10.38
2.0141  1.0073  3.0160 u  0.0054 u  5.0 MeV
10.39
2  3 : 0160  4.0026  2  1.0073 u  0.0148 u  13.8 MeV
The conversion from mass (amu) to energy (MeV) is just multiplying by c2 . To compute the net Q per
4
2
He produced, note that
the reactions (10.37) and (10.38) must each proceed twice for every production of helium in (10.39). Also, the two positrons thus
produced in (10.37) will immediately annihilate with two electrons and produce 4×511 keV = 2:0 MeV. The net Q is thus 2 × 0.5
+ 2× 5.0 + 13.8 + 2:0 = 26.8 MeV per 42 He.
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